What is the arc length of #f(x) = -cscx # on #x in [pi/12,(pi)/8] #?

1 Answer
Jun 2, 2018

#L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oosum_(k=0)^(2n-1)((1/2),(n))((2n-1),(k))(-1)^kint_(pi/12)^(pi/8)sec^(2(n-k)-1)xdx# units.

Explanation:

#f(x)=-cscx#

#f'(x)=cscxcotx#

Arc length is given by:

#L=int_(pi/12)^(pi/8)sqrt(1+csc^2xcot^2x)dx#

Rearrange:

#L=int_(pi/12)^(pi/8)cscxcotxsqrt(1+sin^2xtan^2x)dx#

For #x in [pi/12,pi/8]#, #sin^2xtan^2x<1#. Take the series expansion of the square root:

#L=int_(pi/12)^(pi/8)cscxcotx{sum_(n=0)^oo((1/2),(n))(sinxtanx)^(2n)}dx#

Isolate the #n=0# term:

#L=int_(pi/12)^(pi/8)cscxcotxdx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(sinxtanx)^(2n-1)dx#

Rearrange:

#L=[-cscx]_ (pi/12)^(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)(secx-1/secx)^(2n-1)dx#

Apply binominal expansion:

#L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^(pi/8)sum_(k=0)^(2n-1)((2n-1),(k))(secx)^(2n-1-k)(-1/secx)^kdx#

Rearrange:

#L=csc(pi/12)-csc(pi/8)+sum_(n=1)^oosum_(k=0)^(2n-1)((1/2),(n))((2n-1),(k))(-1)^kint_(pi/12)^(pi/8)sec^(2(n-k)-1)xdx#